The kinetic energy of a particle executing $S.H.M.$ is $16 \, J$ when it is at its mean position. If the mass of the particle is $0.32 \, kg$,then what is the maximum velocity of the particle in $m/s$?

$A$ body of mass $5\, g$ is executing $S.H.M.$ about a point with amplitude $10\, cm$. Its maximum velocity is $100\, cm/s$. Its velocity will be $50\, cm/s$ at a distance of:

$A$ particle executes simple harmonic motion with amplitude $A$ and period $T$. If it is halfway between the mean position and the extreme position,then its speed at that point is:

The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $x/7 \text{ s}$. The value of $x$ is (Take $\pi = 22/7$)

The angular velocities of three bodies in simple harmonic motion are ${\omega _1}, {\omega _2}, {\omega _3}$ with their respective amplitudes as ${A_1}, {A_2}, {A_3}$. If all the three bodies have the same maximum velocity,then:

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively $5 \,m \,s^{-1}$ and $10 \,m \,s^{-2}$, then the time period of oscillation of the particle is